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LOGARITHMS (2)LEARNING OUTCOMES AND ASSESSMENT STANDARDS

Learning outcome 1: Number and number relationshipsAssessment Standard 12.1.2

Demonstrates an understanding of the definition of a logarithm and any laws needed to solve real life problems (eg Growth and decay).

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OverviewIn this lesson you will:

learn how to change the base of a logarithm

use the change of base law to simplify expressions and solve equation

use your calculator correctly

solve real life problems.

Lesson

Log equationsConverting to exponential form

Solve for x: log2 x = 3

Solution

Restriction on x is x ≥ 0

So: 23 = x

x = 8

Dropping the logarithm both sides

Solve for x: log2 (x − 2) + log

2 (x − 3) = 1

Solution

Restriction on x is

x – 2 > 0 → x > 2

x – 3 > 0 → x > 3

Thus: x > 3

log2 (x − 2)(x − 3) = 1 Change 1 into log

22

log2 (x − 2)(x − 3) = log

22 Both sides now have log

2 …

x2 − 5x + 6 = 2 Solve by dropping logs

x2 − 5x + 6 = 2

x2 − 5x + 4 = 0

(x − 4)(x − 1) = 0

x = 4 or x = 1

x = 4 only since x > 3

Converting to log form:

Solve for x: 5x = 9

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Lesson

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Solution

x = log 9

_ log 5

(def: if ax = b then x = log b

_ log a

) Use your calculator

x = 1,365

Change of base lawlog

m x =

loga x _

loga m

Examples

1. Evaluate: lo g 1 _ 2 8 − log

9 27

= log 8

_ log 1 _ 2

+ log 72

_ log 9

= log 23

_ log 2–1 +

log 33

_ log 32

= 3log 2

_ –1log 2

+ 3log 3

_ 2log 3

= –3 + 3 _ 2

= – 3 _ 2

Solution

Prime factors

2. If logx 2 = log

y 64, show that y = x6

Solution

log 2

_ log x

= log 64

_ log y

Put numbers on one side

log y

_ log x

= log 26

_ log 2

log y

_ log x

= 6

log y = 6 log x

log y = log x6

y = x6

Use your calculator to evaluate log2 3

(Use the log button)

Real life applications1. The radioactive decay of a certain substance is given by the formula

m(t) = 500(0,92)t where m is the mass of radioactive substance in grams and t is the number of years.

a) What was the initial mass of the radioactive material?

Solution

Make t = 0

m(0) = 500(0,92)0

= 500 gms

b) By what percentage does the radioactivity decrease by each year?

Solution

(1 − i)1 = 0,92

∴ –i = –0,08

∴ r = 8%

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c) Determine the mass of the substance after 50 years

Solution

m(50) = 500(0,92)50

= 7,73 gms

d) After how many years will the substance’s mass be less than 1 gram?

Solution

500(0,92)n < 1

(0,92)n < 1 _ 500

n > log 1 _

500 _

log 0,92 since log 0,92 and log 1 _ 500 are both negative

n > 74,5

n = 75

Remember: the log of any number between 1 and 0 is always negative. When we have an inequality and we divide by a negative, the sign must swap around.

Activity 1The Richter Scale was developed in 1935 by American scientist Charles F Richter (1900 to 1985). It is used to measure the magnitude of an earthquake by taking the logarithm of the amplitude of waves recorded by a seismograph. The magnitude of an earthquake is defined by M = log I _ s , where I is the intensity of the earthquake (measured by the amplitude of a seismograph reading taken 100 km from the epicentre of the earthquake) and S is the intensity of a “standard” earthquake (the amplitude of which is 1 micron = 10−4 cm).

a) Calculate the magnitude of a standard earthquake.

b) If the intensity of the Ceres earthquake was x micron, express log S in terms of x and the magnitude of the Ceres earthquake.

c) If the intensity of the Sumatra earthquake was y micron, express log S in terms of y and the magnitude of the Sumatra earthquake.

d) Determine the value of y _ x . Explain what this value means.

Activity 21. Calculate correctly to three decimal places:

log3 8

2. Solve for x: 14x = log 2,4 3. Solve: 3x = 5

4. Solve for x: 22x − 11.2x + 24 = 0

Activity 31. Solve for x:

a) 2x + 1 = 3x − 2 b) 7x = 0,5 c) 7x + 7x + 2 = 250

2. If 3x = 6 prove (no calculator) that x = 1 + log3 2.

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ANSWERS AND ASSESSMENT

Lesson 7

Activity 12. log

5 30

_ 6 = log

55 = 1

3. log2 83 − log

2 16

= log2 (8·8·8)

_ 8·2

= log2 32

= log2 25

= 5 log2 2 = 5

4. log 25 + log 8 − log 2 = log ( 25 × 8 _

2 )

= log 100 = log 102 = 2log 10 = 2

5. 1 _ 2 log 25 − log 2 + 1 _ 3 log 64 = 1 _ 2 log (5)2 − log 2 + 1 _ 3 log (2)6 {prime factorise}

= log 5 − log 2 + 2 log 2 = log 5 + log 2 = log (2.5) {law 1} = log 10 = 1

Activity 21. log

5 125 _ 625

= log5 1 _ 5 = log

5 5−1 = − 1

2. log

5 125 _

log5 625

= log

5 53

_ log

5 54 =

3log5 5 _

4log5 5

= 3 _ 4

3. log3 (81 × 9) = log

3 36 = 6

4. log

3 81 _

log3 3–2 =

4log3 3 _

–2log3 3

= –2

5. log 64 + log 2

__ log 32 − log 2 = log 26 + log 2

__ log 25 − log 2

= 6 log 2 + log 2

__ 5 log 2 − log 2

= 7 log 2

_ 4 log 2

= 7 _ 4

6. log9(log

2 8) = log 9(log

223)

= log 9 (3 log

2 2)

= log9 3

= log 3

_ log 9

= log 3

_ 2 log 3

= 1 _ 2

7. 3log 5 – 6log 2

__ 2log 5 – 4log 2

= 3(log 5 – 2log 2)

__ 2(log 5 – 2log 2)

= 3 _ 2

10. (3)(4) = 12

13. log4 ( 1 _ 12 × 1 _

6 × 18 _ 1 ) = log

4 1 _ 4 = log

4 4–1 = −1

15. 3log3 3 + log

2 2−3 + log

7 72

= 3 log3 3 − 3 log

2 2 + 2 log

7 7

= 3 − 3 + 2 = 2

16. log3(33 )

1 _ 4 + log

3 3

1 _ 2 = 3 _ 4 log

3 3 + 1 _ 2 log

3 3 = 5 _ 4

19. log6 (25 )

3 _ 5 + log

6 (33 )

2 _ 3 + log

6 (52 )

1 _ 2 – log

6 10

= log6 (8 × 9 × 5)

__ 10 = log6 36 = log

6 62 = 2

Lesson 8

Activity 1 a) Calculate the magnitude of a standard earthquake.

M = log I _ S

I = 10−4 S = 10−4

M = log 10–4

_ 10–4

M = log 1

m = 0

b) If the intensity of the Ceres earthquake was x micron, express

log S in terms of x and the magnitude of the Ceres earthquake.

M = x _ S

6,3 = log x − log s

log S = 6,3 − log x

c) If the intensity of the Sumatra earthquake was y micron, express

log S in terms of y and the magnitude of the Sumatra earthquake.

9,1 = log y _ S

9,1 = log y − log S

log S = log y − 9,1

d) Determine the value of y _ x . Explain what this value

means.

log x − 6,3 = log y − 9,1

2,8 = log y − log x

2,8 = log y _ x 102,8 =

y _ x

y _ x = 631

Intensity of the Sumatra earthquake 631 times that of the Ceres earthquake.

Activity 41. log

3 8 =

log 8 _

log 3

= 1,893 [keys 8 log ÷ 3 log = ]

14x = log 2,4

2. 14x = (log 2,4)

x = log (log 2,4)

__ log 14

= –0.366

3. log 3x = log 5

x = log 5

_ log 3

= 1,465

4. let 2x = k

k2 − 11k + 24 = 0

(k − 3)(k − 8) = 0

∴ 2x = 3 or 2x = 8

x = log 3

_ log 2

or 2x = 23

x = 1,585 or x = 3

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Activity 51. a) 2x+1 = 3x−2

2·2x = 3x·3–2

∴ ( 2 _ 3 ) x

= 1 _ 18

∴ x = log ( 1 _ 18 )

_ log ( 2 _ 3 )

∴ x = 7,12

b) 7x = 0,5

x = log 0,5

_ log 7

= −0, 36

c) 7x + 7x·72 = 250

let 7x = k

50k = 250

k = 5

7x = 5

x = log 5

_ log 7

x = 0,83

2. x = log 6

_ log 3

= log 3 + log 2

__ log 3

= 1 + log3 2

TIPS FOR THE TEACHER

Lesson 7Logs are easier that they were in the previous syllabus.

A lot of emphasis must be placed on using logs to solve equations and real-life applications.

Lesson 8Logs are easier than they were in the previous syllabus.

A lot of emphasis must be placed on using logs to solve equations and real life applications.

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